Does anyone know how to calculate the pointwise limit of fn on [-1,1]

of fn:[-1,1] -> R: x ->fn(x) = (x)/(1+(n^2)(x^2))

Do we take the limit as n-> infinity

and then sub in [-1,1] or what do we do with these numbers.

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- May 16th 2010, 11:39 AMfrogPointwise LimitDoes anyone know how to calculate the pointwise limit of fn on [-1,1]

of fn:[-1,1] -> R: x ->fn(x) = (x)/(1+(n^2)(x^2))

Do we take the limit as n-> infinity

and then sub in [-1,1] or what do we do with these numbers.

- May 16th 2010, 11:41 AMroninpro
A pointwise limit simply means that you fix $\displaystyle x\in [-1,1]$ and then send $\displaystyle n\to \infty$.

Can you see what happens? - May 16th 2010, 11:46 AMfrog
what you sub in [-1,1] first and then get the limit as n-> infinity???

so you get (1)/(1+n^2)

which then goes to zero as n-> infinity - May 16th 2010, 11:55 AMroninpro
You put a point $\displaystyle x$ in, then take the limit. Actually, you have

$\displaystyle \lim_{n\to \infty} \frac{x}{1+n^2x^2}$

but this still goes to 0. (You should probably make sure.) - May 16th 2010, 12:02 PMfrog
But how do you sub in the point? Do you sub in somewhere in the regio of [-1,1] such as zero. Or what?

so it would then be 0/1 which is zero as its lim 0??

Thanks for the help but im still unsure about what exactly you are telling me to do. I have examples that are [0,1] but thought that this one could be a bit different.

Was my answer in the above post write or is this write???

it's zero either way but no marks will be give if the method is wrong.... - May 16th 2010, 12:10 PMroninpro
Your above post was not quite correct.

You have to argue that the limit goes to 0 for any arbitrary point $\displaystyle x$. In other words, fix an arbitrary $\displaystyle x\in [-1, 1]$. Then,

$\displaystyle

\lim_{n\to \infty} \frac{x}{1+n^2x^2}=x\lim_{n\to \infty} \frac{1}{1+n^2x^2}=0

$

since the denominator goes to infinity. - May 16th 2010, 12:13 PMfrog
So the answer is zero? Just like that. Doesn't matter about what x is an element of?

Is this how we always solve it.

i.e take the limit, and whatever value you have for x after this is what you use for uniformity?

So in this case i would use zero?? - May 16th 2010, 12:17 PMroninpro
There's a little bit of a catch in the above. If $\displaystyle x=0$, the denominator doesn't go to infinity, but the limit still goes to 0.

If you want to argue for uniform convergence, in this case, you would use the zero function as the uniform limit. (It's the only possible choice - if a sequence of functions converges uniformly, then it has to converge to the pointwise limit.) - May 16th 2010, 12:23 PMfrog
That's great thanks.